# Proving the chain rule for complex functions

I'm familiar with the Real Analysis proof of the chain rule (i.e. looking at the difference quotient for both $g(f(z))$ and for $f(z)$), and I'm familiar with another proof using the Weierstrass definition of differentiability (differentiable iff there is a continuous function such that ...).

But in Bak and Newman's Complex Analysis they give a hint for proving that the composition of differentiable functions is differentiable.

Begin by noting $$g(f(z+h))-g(f(z)) = [g'(f(z))+\epsilon][f(z+h)-f(z)]$$ where $\epsilon\rightarrow 0$ as $h\rightarrow 0$.

This seems to me to be practically assuming the thing we're trying to prove. What is the justification for this equation? It's not an equation I've encountered in earlier studies--am I supposed to be familiar with it?

This also isn't the first time that I've encountered an expression involving quantities going to 0 like this, which I didn't fully understand (like when reading about Machine Learning or Statistics). Is there a book I can consult to better understand the theory around this? As far as I recall it wasn't in baby Rudin or Ross's Analysis textbook, and yet it comes up kind of often and is thrown around casually.

## 1 Answer

Definitely not a nice way to introduce this (I agree with you), but what they used is the fact that: $g(b)-g(a)=(b-a).g'(a)+o(b-a)$, which is litteraly the definition of the derivative at point a.

Now you can use b = f(z+h), a = f(z), and the previous equation will yield the one given by Bak and Newman.